Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $n = \dfrac{-9k^2 + 9k}{k - 5} \div \dfrac{k^2 + 5k - 6}{k - 5} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $n = \dfrac{-9k^2 + 9k}{k - 5} \times \dfrac{k - 5}{k^2 + 5k - 6} $ First factor the quadratic. $n = \dfrac{-9k^2 + 9k}{k - 5} \times \dfrac{k - 5}{(k - 1)(k + 6)} $ Then factor out any other terms. $n = \dfrac{-9k(k - 1)}{k - 5} \times \dfrac{k - 5}{(k - 1)(k + 6)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ -9k(k - 1) \times (k - 5) } { (k - 5) \times (k - 1)(k + 6) } $ $n = \dfrac{ -9k(k - 1)(k - 5)}{ (k - 5)(k - 1)(k + 6)} $ Notice that $(k - 5)$ and $(k - 1)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ -9k\cancel{(k - 1)}(k - 5)}{ (k - 5)\cancel{(k - 1)}(k + 6)} $ We are dividing by $k - 1$ , so $k - 1 \neq 0$ Therefore, $k \neq 1$ $n = \dfrac{ -9k\cancel{(k - 1)}\cancel{(k - 5)}}{ \cancel{(k - 5)}\cancel{(k - 1)}(k + 6)} $ We are dividing by $k - 5$ , so $k - 5 \neq 0$ Therefore, $k \neq 5$ $n = \dfrac{-9k}{k + 6} ; \space k \neq 1 ; \space k \neq 5 $